A) \[\sqrt{\frac{2G{{M}_{e}}+{{R}_{e}}}{{{R}_{e}}}}\]
B) \[\sqrt{\frac{2G{{M}_{e}}m}{{{R}_{e}}}}\]
C) \[\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\]
D) \[\frac{G{{M}_{e}}}{{{R}_{e}}}\]
Correct Answer: C
Solution :
The work done to take a body of mass m where the earths attraction is negligible, is \[W=\frac{G{{M}_{e}}m}{{{R}_{e}}}\] This is equal to kinetic energy on surface of earth. i.e., \[\frac{1}{2}m{{v}_{e}}^{2}=\frac{G{{M}_{e}}m}{{{R}_{e}}}\] \[\Rightarrow \] \[{{v}_{e}}\sqrt{\frac{2G{{M}_{e}}}{{{R}_{e}}}}\]
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